The equation you set up will work just fine. I think the best way to solve this using your equation is to just estimate √2 as 1.4.

x + x + 1.4x = 16 + 16(1.4)

3.4x = 38.4

x = ~11.3


The hypotenuse would be 1.4 (11.3) = ~15.82

It would be pretty clear that the answer would be 16.

Also you could do the following:

Assume that the hypotenuse is x and the sides of the triangle are x√2/2.

(x√2/2)+(x√2/2)+x=16 + 16√2
2x√2/2 + x = 16 + 16√2
x√2 + x = 16 + 16√2

x =16
x√2=16√2

damrkstr wrote:The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!

taking ahead your approach...
we can see
x + x + x√2 = 16 + 16√2
pick numbers...
first take x=8, it gives
8+8+8√2 which is not equal to 16+16√2. hence xis not eual to 8.
now pick x=8√2
8√2+8√2+16 = 16+16√2

hence the side of the triangle is 8√2 and the hypotenuse is 16.
U arrived at the equation correctly but then had to just pick numbers.
hope it helps...

damrkstr wrote:The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!

one more simple and elegant approach.....
x+x+x√2= 16 + 16√2
2x+x√2=16(1+√2)
x√2(√2+1)=16(1+√2)
which gives x√2 = 16 or x = 8√2

2x²=y² and 2x+y=16+16√2

y=16[/list]

jimmiejaz wrote:

damrkstr wrote:The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!

one more simple and elegant approach.....
x+x+x√2= 16 + 16√2
2x+x√2=16(1+√2)
x√2(√2+1)=16(1+√2)
which gives x√2 = 16 or x = 8√2

x = 8√2 is not 16 and also, how did you turn 2x+x√2 into x√2(√2+1)?

cartera wrote:

jimmiejaz wrote:

damrkstr wrote:The perimeter of a certain isosceles right triangle is 16 + 16√2.
What is the length of the hypotenuse of the triangle?

I tried solving it setting x + x + x√2 = 16 + 16√2 but am not solving it correctly. Could someone help me out with the best way to solve the problem?

The answer was supposed to be 16

Thanks for your help!

one more simple and elegant approach.....
x+x+x√2= 16 + 16√2
2x+x√2=16(1+√2)
x√2(√2+1)=16(1+√2)
which gives x√2 = 16 or x = 8√2

x = 8√2 is not 16 and also, how did you turn 2x+x√2 into x√2(√2+1)?

Hi,
Let me answer the second part of your question:
Remembering that √2*√2=2, then 2x=(√2*√2)x
Accordingly 2x+x√2=(√2*√2)x+x√2=x√2(√2+1)